A ray of light is sent along the line `x-2y+5 = 0 ` upon reaching the line `3x- 2y + 7 = 0 ` the ray is reflected from it . Find the equation of the containing the reflected ray .

Answer

To find the equation of the line containing the reflected ray after a ray of light reflects off the line \(3x - 2y + 7 = 0\), we will follow these steps:

### Step 1: Identify the equations of the incident and reflecting lines.

The incident ray is given by the equation:
\[ x - 2y + 5 = 0 \]
The reflecting line is given by:
\[ 3x - 2y + 7 = 0 \]

### Step 2: Find a point on the incident ray.

To find a point on the line \(x - 2y + 5 = 0\), we can choose \(x = 1\):
\[ 1 - 2y + 5 = 0 \implies 2y = 6 \implies y = 3 \]
Thus, the point on the incident ray is \(P(1, 3)\).

### Step 3: Find the image of point \(P\) with respect to the reflecting line.

To find the image of point \(P(1, 3)\) with respect to the line \(3x - 2y + 7 = 0\), we use the formula for finding the image of a point \((x_1, y_1)\) across the line \(Ax + By + C = 0\):
\[ \text{Image} = \left( x_1 - \frac{2A(Ax_1 + By_1 + C)}{A^2 + B^2}, y_1 - \frac{2B(Ax_1 + By_1 + C)}{A^2 + B^2} \right) \]
Here, \(A = 3\), \(B = -2\), and \(C = 7\).

Calculating \(Ax_1 + By_1 + C\):
\[ 3(1) - 2(3) + 7 = 3 - 6 + 7 = 4 \]
Now, substituting into the image formula:
\[ \text{Image} = \left( 1 - \frac{2 \cdot 3 \cdot 4}{3^2 + (-2)^2}, 3 - \frac{2 \cdot (-2) \cdot 4}{3^2 + (-2)^2} \right) \]
Calculating \(A^2 + B^2\):
\[ 3^2 + (-2)^2 = 9 + 4 = 13 \]
Now substituting:
\[ \text{Image} = \left( 1 - \frac{24}{13}, 3 + \frac{16}{13} \right) = \left( \frac{13 - 24}{13}, \frac{39 + 16}{13} \right) = \left( -\frac{11}{13}, \frac{55}{13} \right) \]

### Step 4: Find the equation of the reflected ray.

The reflected ray will pass through the point \(P(1, 3)\) and the image point \(\left(-\frac{11}{13}, \frac{55}{13}\right)\). We can find the slope \(m\) of the line through these two points:
\[ m = \frac{\frac{55}{13} - 3}{-\frac{11}{13} - 1} = \frac{\frac{55}{13} - \frac{39}{13}}{-\frac{11}{13} - \frac{13}{13}} = \frac{\frac{16}{13}}{-\frac{24}{13}} = -\frac{2}{3} \]

Using point-slope form \(y - y_1 = m(x - x_1)\) with point \(P(1, 3)\):
\[ y - 3 = -\frac{2}{3}(x - 1) \]
Multiplying through by 3 to eliminate the fraction:
\[ 3(y - 3) = -2(x - 1) \implies 3y - 9 = -2x + 2 \implies 2x + 3y - 11 = 0 \]

### Final Answer

Thus, the equation of the line containing the reflected ray is:
\[ 2x + 3y - 11 = 0 \]